What is the integral of 1/(6x^2+7x+2) ?

The integral of 1/(6x^2+7x+2) is -ln|12x+8|+ln|12x+6|+C

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integral of 1/(6x^2+7x+2)

\int \frac{1}{6 x ^{2} + 7 x + 2} d x

- ln ∣ 12 x + 8 ∣ + ln ∣ 12 x + 6 ∣ + C

Solution steps

\int \frac{1}{6 x ^{2} + 7 x + 2} d x

Complete the square

6 x^{2} + 7 x + 2 : 6 (x + \frac{7}{12})^{2} - \frac{1}{24}

= \int \frac{1}{6 ( x + \frac{7}{12} ) ^{2} - \frac{1}{24}} d x

Apply u-substitution

: \int \frac{24}{144 u ^{2} - 1} d u

= \int \frac{24}{144 u ^{2} - 1} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 24 \cdot \int \frac{1}{144 u ^{2} - 1} d u

Factor

144 u^{2} - 1 : - (- 144 u^{2} + 1)

= 24 \cdot \int \frac{1}{- ( - 144 u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 24 (- \int \frac{1}{- 144 u ^{2} + 1} d u)

Apply Integral Substitution

= 24 (- \int \frac{1}{12 ( - v ^{2} + 1 )} d v)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 24 (- \frac{1}{12} \cdot \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 24 (- \frac{1}{12} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 24 (- \frac{1}{12} (\frac{ln 12 ( x + \frac{7}{12} ) + 1}{2} - \frac{ln 12 ( x + \frac{7}{12} ) - 1}{2}))

Simplify

24 (- \frac{1}{12} (\frac{ln 12 ( x + \frac{7}{12} ) + 1}{2} - \frac{ln 12 ( x + \frac{7}{12} ) - 1}{2})) : - ln ∣ 12 x + 8 ∣ + ln ∣ 12 x + 6 ∣

= - ln ∣ 12 x + 8 ∣ + ln ∣ 12 x + 6 ∣

Add a constant to the solution

= - ln ∣ 12 x + 8 ∣ + ln ∣ 12 x + 6 ∣ + C

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What is the integral of 1/(6x^2+7x+2) ?
The integral of 1/(6x^2+7x+2) is -ln|12x+8|+ln|12x+6|+C